<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Verify that $\frac{2}{{{r}^{2}}-1}=\frac{1}{r-1}-\frac{1}{r+1}$, where $r\in \mathbb{Z}$, $r\ge 2$.</span></div><p style="text-align: right;">[1]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence evaluate $\sum\limits_{r=2}^{n}{\frac{1}{{{r}^{2}}-1}}$, where $n\in \mathbb{Z}$, $n\ge 2$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Using the result in part (ii), show that $\sum\limits_{r=2}^{N}{\frac{1}{{{\left( r+1 \right)}^{2}}}<\frac{3}{4}}$ for all $N\in \mathbb{Z}$, $N\ge 2$.</span></div><p style="text-align: right;">[3]</p></li></ol>

Verify that $\frac{2}{{{r}^{2}}-1}=\frac{1}{r-1}-\frac{1}{r+1}$, where $r\in \mathbb{Z}$, $r\ge 2$.

Hence evaluate $\sum\limits_{r=2}^{n}{\frac{1}{{{r}^{2}}-1}}$, where $n\in \mathbb{Z}$, $n\ge 2$.

Using the result in part (ii), show that $\sum\limits_{r=2}^{N}{\frac{1}{{{\left( r+1 \right)}^{2}}}<\frac{3}{4}}$ for all $N\in \mathbb{Z}$, $N\ge 2$.

2020 HCI Promo Q6

2020 HCI Promo Q6

Video Solution:

Solution:

Solution locked